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Proof step 26 from proof size-two (previous step)

s0 ++ a = ∅ ++ a0 ++ b ⊦ 1 = # s0, a0 = b, s0 = ∅ ++ b, a ∈ s0, ∃ a. s0 = ∅ ++ a

User applied rule exists right

And obtained the following new goals

Proof step 27:
s0 ++ a = ∅ ++ a0 ++ b ⊦ 1 = # s0, a0 = b, s0 = ∅ ++ b, s0 = ∅ ++ a0, a ∈ s0, ∃ a. s0 = ∅ ++ a

Rule argument

pos: right 5 terms: a0 abort: F discard: F computed: F

Additional information

Used terms: a0